9.E: Hypothesis Testing with One Sample (Exercises)
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These are homework exercises to accompany the Textmap created for
"Introductory Statistics" by OpenStax.
9.1: Introduction
9.2: Null and Alternative Hypotheses
Q 9.2.1
Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, \(H_{0}\), and the alternative hypothesis. \(H_{a}\), in terms of the appropriate parameter \((\mu \text{or} p)\).
- The mean number of years Americans work before retiring is 34.
- At most 60% of Americans vote in presidential elections.
- The mean starting salary for San Jose State University graduates is at least $100,000 per year.
- Twenty-nine percent of high school seniors get drunk each month.
- Fewer than 5% of adults ride the bus to work in Los Angeles.
- The mean number of cars a person owns in her lifetime is not more than ten.
- About half of Americans prefer to live away from cities, given the choice.
- Europeans have a mean paid vacation each year of six weeks.
- The chance of developing breast cancer is under 11% for women.
- Private universities' mean tuition cost is more than $20,000 per year.
S 9.2.1
- \(H_{0}: \mu = 34; H_{a}: \mu \neq 34\)
- \(H_{0}: p \leq 0.60; H_{a}: p > 0.60\)
- \(H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000\)
- \(H_{0}: p = 0.29; H_{a}: p \neq 0.29\)
- \(H_{0}: p = 0.05; H_{a}: p < 0.05\)
- \(H_{0}: \mu \leq 10; H_{a}: \mu > 10\)
- \(H_{0}: p = 0.50; H_{a}: p \neq 0.50\)
- \(H_{0}: \mu = 6; H_{a}: \mu \neq 6\)
- \(H_{0}: p ≥ 0.11; H_{a}: p < 0.11\)
- \(H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000\)
Q 9.2.2
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:
- \(p < 0.30\)
- \(p \leq 0.30\)
- \(p \geq 0.30\)
- \(p > 0.30\)
Q 9.2.3
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:
- \(p = 0.20\)
- \(p > 0.20\)
- \(p < 0.20\)
- \(p \leq 0.20\)
S 9.2.3
c
Q 9.2.4
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:
- \(H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5\)
- \(H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5\)
- \(H_{0}: \mu = 4.75, H_{a}: \mu > 4.75\)
- \(H_{0}: \mu = 4.5, H_{a}: \mu > 4.5\)
9.3: Outcomes and the Type I and Type II Errors
Q 9.3.1
State the Type I and Type II errors in complete sentences given the following statements.
- The mean number of years Americans work before retiring is 34.
- At most 60% of Americans vote in presidential elections.
- The mean starting salary for San Jose State University graduates is at least $100,000 per year.
- Twenty-nine percent of high school seniors get drunk each month.
- Fewer than 5% of adults ride the bus to work in Los Angeles.
- The mean number of cars a person owns in his or her lifetime is not more than ten.
- About half of Americans prefer to live away from cities, given the choice.
- Europeans have a mean paid vacation each year of six weeks.
- The chance of developing breast cancer is under 11% for women.
- Private universities mean tuition cost is more than $20,000 per year.
S 9.3.1
- Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
- Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
- Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
- Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
- Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
- Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
- Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
- Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
- Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
- Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.
Q 9.3.2
For statements a-j in Exercise 9.109 , answer the following in complete sentences.
- State a consequence of committing a Type I error.
- State a consequence of committing a Type II error.
Q 9.3.3
When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?
- To conclude the drug is safe when in, fact, it is unsafe.
- Not to conclude the drug is safe when, in fact, it is safe.
- To conclude the drug is safe when, in fact, it is safe.
- Not to conclude the drug is unsafe when, in fact, it is unsafe.
S 9.3.3
b
Q 9.3.4
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.
- at least 20%, when in fact, it is less than 20%.
- 20%, when in fact, it is 20%.
- less than 20%, when in fact, it is at least 20%.
- less than 20%, when in fact, it is less than 20%.
Q 9.3.4
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?
The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours
- is more than seven hours.
- is at most seven hours.
- is at least seven hours.
- is less than seven hours.
S 9.3.4
d
Q 9.3.5
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:
- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
- to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
- to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher
9.4: Distribution Needed for Hypothesis Testing
Q 9.4.1
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is \(\bar{X} \sim\) ________________
- \(N\left(7.24, \frac{1.93}{\sqrt{22}}\right)\)
- \(N\left(7.24, 1.93\right)\)
- \(t_{22}\)
- \(t_{21}\)
S 9.4.1
d
9.5: Rare Events, the Sample, Decision and Conclusion
Q 9.5.1
The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.
- Is this a test of one mean or proportion?
-
State the null and alternative hypotheses.
\(H_{0}\) : ____________________ \(H_{a}\) : ____________________ - Is this a right-tailed, left-tailed, or two-tailed test?
- What symbol represents the random variable for this test?
- In words, define the random variable for this test.
-
Calculate the following:
- \(x =\) ________________
- \(n =\) ________________
- \(p′ =\) _____________
- Calculate \(\sigma_{x} =\) __________. Show the formula set-up.
- State the distribution to use for the hypothesis test.
- Find the \(p\text{-value}\).
-
At a pre-conceived \(\alpha = 0.05\), what is your:
- Decision:
- Reason for the decision:
- Conclusion (write out in a complete sentence):
9.6: Additional Information and Full Hypothesis Test Examples
For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.
Note
If you are using a Student's \(t\) - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)
Q 9.6.1.
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using \(\alpha = 0.05\), is the data highly inconsistent with the claim?
S 9.6.1
- \(H_{0}: \mu \geq 50,000\)
- \(H_{a}: \mu < 50,000\)
- Let \(\bar{X} =\) the average lifespan of a brand of tires.
- normal distribution
- \(z = -2.315\)
- \(p\text{-value} = 0.0103\)
- Check student’s solution.
-
- alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is less than 0.05.
- Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
- \((43,537, 49,463)\)
Q 9.6.2
From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
Q 9.6.3
The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?
S 9.6.3
- \(H_{0}: \mu = $1.00\)
- \(H_{a}: \mu \neq $1.00\)
- Let \(\bar{X} =\) the average cost of a daily newspaper.
- normal distribution
- \(z = –0.866\)
- \(p\text{-value} = 0.3865\)
- Check student’s solution.
-
- \(\alpha: 0.01\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.01.
- Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
- \(($0.84, $1.06)\)
Q 9.6.4
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
Q 9.6.5
The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let \(x =\) the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?
S 9.6.5
- \(H_{0}: \mu = 10\)
- \(H_{a}: \mu \neq 10\)
- Let \(\bar{X}\) the mean number of sick days an employee takes per year.
- Student’s t -distribution
- \(t = –1.12\)
- \(p\text{-value} = 0.300\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
- \((4.9443, 11.806)\)
Q 9.6.6
In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?
Q 9.6.7
Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?
S 9.6.7
- \(H_{0}: p \geq 0.6\)
- \(H_{a}: p < 0.6\)
- Let \(P′ =\) the proportion of students who feel more enriched as a result of taking Elementary Statistics.
- normal for a single proportion
- 1.12
- \(p\text{-value} = 0.1308\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.
-
Confidence Interval: \((0.409, 0.654)\)
The “plus-4s” confidence interval is \((0.411, 0.648)\)
Q 9.6.8
A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.
Q 9.6.9
Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.
S 9.6.9
- \(H_{0}: \mu = 4\)
- \(H_{a}: \mu \neq 4\)
- Let \(\bar{X}\) the average I.Q. of a set of brown trout.
- two-tailed Student's t-test
- \(t = 1.95\)
- \(p\text{-value} = 0.076\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05
- Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
- \((3.8865,5.9468)\)
Q 9.6.10
According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?
Q 9.6.11
A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.
S 9.6.11
-
\(H_{0}: p \geq 0.13\)
- \(H_{a}: p < 0.13\)
- Let \(P′ =\) the proportion of Americans who have seen or sensed angels
- normal for a single proportion
- –2.688
- \(p\text{-value} = 0.0036\)
- Check student’s solution.
-
- alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\)e is less than 0.05.
- Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.
-
\((0, 0.0623)\).
The“plus-4s” confidence interval is (0.0022, 0.0978)
Q 9.6.12
The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?
Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
Q 9.6.13
Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.
S 9.6.13
- \(H_{0}: \mu \geq 129\)
- \(H_{a}: \mu < 129\)
- Let \(\bar{X} =\) the average time in seconds that Terri finishes Lap 4.
- Student's t -distribution
- \(t = 1.209\)
- 0.8792
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
- \((128.63, 130.37)\)
Q 9.6.14
Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.
Note
The following questions were written by past students. They are excellent problems!
Q 9.6.15
"Asian Family Reunion," by Chau Nguyen
Every two years it comes around.
We all get together from different towns.
In my honest opinion,
It's not a typical family reunion.
Not forty, or fifty, or sixty,
But how about seventy companions!
The kids would play, scream, and shout
One minute they're happy, another they'll pout.
The teenagers would look, stare, and compare
From how they look to what they wear.
The men would chat about their business
That they make more, but never less.
Money is always their subject
And there's always talk of more new projects.
The women get tired from all of the chats
They head to the kitchen to set out the mats.
Some would sit and some would stand
Eating and talking with plates in their hands.
Then come the games and the songs
And suddenly, everyone gets along!
With all that laughter, it's sad to say
That it always ends in the same old way.
They hug and kiss and say "good-bye"
And then they all begin to cry!
I say that 60 percent shed their tears
But my mom counted 35 people this year.
She said that boys and men will always have their pride,
So we won't ever see them cry.
I myself don't think she's correct,
So could you please try this problem to see if you object?
S 9.6.15
- \(H_{0}: p = 0.60\)
- \(H_{a}: p < 0.60\)
- Let \(P′ =\) the proportion of family members who shed tears at a reunion.
- normal for a single proportion
- –1.71
- 0.0438
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis.
- Reason for decision: \(p\text{-value} < \alpha\)
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the \(p\text{-value}\) and alpha are quite close, so other tests should be done.
- We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. \((0.3829, 0.6171)\). The“plus-4s” confidence interval (see chapter 8) is \((0.3861, 0.6139)\)
Note that here the “large-sample” \(1 - \text{PropZTest}\) provides the approximate \(p\text{-value}\) of 0.0438. Whenever a \(p\text{-value}\) based on a normal approximation is close to the level of significance, the exact \(p\text{-value}\) based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.
Q 9.6.16
"The Problem with Angels," by Cyndy Dowling
Although this problem is wholly mine,
The catalyst came from the magazine, Time.
On the magazine cover I did find
The realm of angels tickling my mind.
Inside, 69% I found to be
In angels, Americans do believe.
Then, it was time to rise to the task,
Ninety-five high school and college students I did ask.
Viewing all as one group,
Random sampling to get the scoop.
So, I asked each to be true,
"Do you believe in angels?" Tell me, do!
Hypothesizing at the start,
Totally believing in my heart
That the proportion who said yes
Would be equal on this test.
Lo and behold, seventy-three did arrive,
Out of the sample of ninety-five.
Now your job has just begun,
Solve this problem and have some fun.
Q 9.6.17
"Blowing Bubbles," by Sondra Prull
Studying stats just made me tense,
I had to find some sane defense.
Some light and lifting simple play
To float my math anxiety away.
Blowing bubbles lifts me high
Takes my troubles to the sky.
POIK! They're gone, with all my stress
Bubble therapy is the best.
The label said each time I blew
The average number of bubbles would be at least 22.
I blew and blew and this I found
From 64 blows, they all are round!
But the number of bubbles in 64 blows
Varied widely, this I know.
20 per blow became the mean
They deviated by 6, and not 16.
From counting bubbles, I sure did relax
But now I give to you your task.
Was 22 a reasonable guess?
Find the answer and pass this test!
S 9.6.17
- \(H_{0}: \mu \geq 22\)
- \(H_{a}: \mu < 22\)
- Let \(\bar{X} =\) the mean number of bubbles per blow.
- Student's t -distribution
- –2.667
- \(p\text{-value} = 0.00486\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is less than 0.05.
- Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
- \((18.501, 21.499)\)
Q 9.6.18
"Dalmatian Darnation," by Kathy Sparling
A greedy dog breeder named Spreckles
Bred puppies with numerous freckles
The Dalmatians he sought
Possessed spot upon spot
The more spots, he thought, the more shekels.
His competitors did not agree
That freckles would increase the fee.
They said, “Spots are quite nice
But they don't affect price;
One should breed for improved pedigree.”
The breeders decided to prove
This strategy was a wrong move.
Breeding only for spots
Would wreak havoc, they thought.
His theory they want to disprove.
They proposed a contest to Spreckles
Comparing dog prices to freckles.
In records they looked up
One hundred one pups:
Dalmatians that fetched the most shekels.
They asked Mr. Spreckles to name
An average spot count he'd claim
To bring in big bucks.
Said Spreckles, “Well, shucks,
It's for one hundred one that I aim.”
Said an amateur statistician
Who wanted to help with this mission.
“Twenty-one for the sample
Standard deviation's ample:
They examined one hundred and one
Dalmatians that fetched a good sum.
They counted each spot,
Mark, freckle and dot
And tallied up every one.
Instead of one hundred one spots
They averaged ninety six dots
Can they muzzle Spreckles’
Obsession with freckles
Based on all the dog data they've got?
Q 9.6.19
"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall
As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.
One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:
Price per box of Mac and Cheese:
- 5 stores @ $2.02
- 15 stores @ $0.25
- 3 stores @ $1.29
- 6 stores @ $0.35
- 4 stores @ $2.27
- 7 stores @ $1.50
- 5 stores @ $1.89
- 8 stores @ 0.75.
I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)
S 9.6.19
- \(H_{0}: \mu \leq 1\)
- \(H_{a}: \mu > 1\)
- Let \(\bar{X} =\) the mean cost in dollars of macaroni and cheese in a certain town.
- Student's \(t\)-distribution
- \(t = 0.340\)
- \(p\text{-value} = 0.36756\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05
- Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
- \((0.8291, 1.241)\)
Q 9.6.20
"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi
THE CHARACTERS (in order of appearance):
- HAMLET, Prince of Denmark and student of Statistics
- POLONIUS, Hamlet’s tutor
- HOROTIO, friend to Hamlet and fellow student
Scene: The great library of the castle, in which Hamlet does his lessons
Act I
(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)
POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!
HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.
POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.
HORATIO (to Hamlet): What should we do, my Lord?
HAMLET: Go to thy purpose, Horatio.
HORATIO: To what end, my Lord?
HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.
(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)
Act II
(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)
POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?
HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.
POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)
HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)
(Curtain falls)
Q 9.6.21
"Untitled," by Stephen Chen
I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.
So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?
S 9.6.21
- \(H_{0}: p = 0.01\)
- \(H_{a}: p > 0.01\)
- Let \(P′ =\) the proportion of errors generated
- Normal for a single proportion
- 2.13
- 0.0165
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis
- Reason for decision: The \(p\text{-value}\) is less than 0.05.
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.
-
Confidence interval: \((0, 0.094)\).
The“plus-4s” confidence interval is \((0.004, 0.144)\).
Q 9.6.22
"Japanese Girls’ Names"
by Kumi Furuichi
It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.
However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.
I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.
Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.
Q 9.6.23
"Phillip’s Wish," by Suzanne Osorio
My nephew likes to play
Chasing the girls makes his day.
He asked his mother
If it is okay
To get his ear pierced.
She said, “No way!”

He argued his point quite well,
Says even my macho pal, Mel,
Has gotten this done.
It’s all just for fun.

Again Phillip complained to his mother,
Saying half his friends (including their brothers)
Are piercing their ears
And they have no fears
He wants to be like the others.
She said, “I think it’s much less.
We must do a hypothesis test.
And if you are right,
I won’t put up a fight.
But, if not, then my case will rest.”
We proceeded to call fifty guys
To see whose prediction would fly.
Nineteen of the fifty
Said piercing was nifty
And earrings they’d occasionally buy.
Then there’s the other thirty-one,
Who said they’d never have this done.
So now this poem’s finished.
Will his hopes be diminished,
Or will my nephew have his fun?
S 9.6.23
- \(H_{0}: p = 0.50\)
- \(H_{a}: p < 0.50\)
- Let \(P′ =\) the proportion of friends that has a pierced ear.
- normal for a single proportion
- –1.70
- \(p\text{-value} = 0.0448\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis
- Reason for decision: The \(p\text{-value}\) is less than 0.05. (However, they are very close.)
- Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
- Confidence Interval: \((0.245, 0.515)\): The “plus-4s” confidence interval is \((0.259, 0.519)\).
Q 9.6.24
"The Craven," by Mark Salangsang
Once upon a morning dreary
In stats class I was weak and weary.
Pondering over last night’s homework
Whose answers were now on the board
This I did and nothing more.
While I nodded nearly napping
Suddenly, there came a tapping.
As someone gently rapping,
Rapping my head as I snore.
Quoth the teacher, “Sleep no more.”
“In every class you fall asleep,”
The teacher said, his voice was deep.
“So a tally I’ve begun to keep
Of every class you nap and snore.
The percentage being forty-four.”
“My dear teacher I must confess,
While sleeping is what I do best.
The percentage, I think, must be less,
A percentage less than forty-four.”
This I said and nothing more.
“We’ll see,” he said and walked away,
And fifty classes from that day
He counted till the month of May
The classes in which I napped and snored.
The number he found was twenty-four.
At a significance level of 0.05,
Please tell me am I still alive?
Or did my grade just take a dive
Plunging down beneath the floor?
Upon thee I hereby implore.
Q 9.6.25
Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.
S 9.6.25
- \(H_{0}: p = 0.40\)
- \(H_{a}: p < 0.40\)
- Let \(P′ =\) the proportion of schoolmates who fear public speaking.
- normal for a single proportion
- –1.01
- \(p\text{-value} = 0.1563\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
- Confidence Interval: \((0.3241, 0.4240)\): The “plus-4s” confidence interval is \((0.3257, 0.4250)\).
Q 9.6.26
Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.
Q 9.6.27
According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.
S 9.6.27
- \(H_{0}: p = 0.14\)
- \(H_{a}: p < 0.14\)
- Let \(P′ =\) the proportion of NYC residents that smoke.
- normal for a single proportion
- –0.2756
- \(p\text{-value} = 0.3914\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
- Confidence Interval: \((0.0502, 0.2070)\): The “plus-4s” confidence interval (see chapter 8) is \((0.0676, 0.2297)\).
Q 9.6.28
The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.
Q 9.6.29
Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.
S 9.6.29
- \(H_{0}: \mu = 69,110\)
- \(H_{0}: \mu > 69,110\)
- Let \(\bar{X} =\) the mean salary in dollars for California registered nurses.
- Student's t -distribution
- \(t = 1.719\)
- \(p\text{-value}: 0.0466\)
- Check student’s solution.
-
- \(\alpha: 0.05\)
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is less than 0.05.
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
- \(($68,757, $73,485)\)
Q 9.6.30
La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.
Q 9.6.31
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?
After conducting the test, your decision and conclusion are
- Reject \(H_{0}\): There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
- Do not reject \(H_{0}\): There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
- Do not reject \(H_{0}\): There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
- Reject \(H_{0}\): There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
S 9.6.31
c
Q 9.6.32
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.
At a 1% level of significance, an appropriate conclusion is:
- There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
- There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
- There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
- There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.
Q 9.6.33
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
At a significance level of \(a = 0.05\), what is the correct conclusion?
- There is enough evidence to conclude that the mean number of hours is more than 4.75
- There is enough evidence to conclude that the mean number of hours is more than 4.5
- There is not enough evidence to conclude that the mean number of hours is more than 4.5
- There is not enough evidence to conclude that the mean number of hours is more than 4.75
S 9.6.33
c
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
State the null and alternate hypothesis.
State the \(p\text{-value}\).
State \(\alpha\).
What is your decision?
Write a conclusion.
Answer any other questions asked in the problem.
Q 9.6.34
According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.
Q 9.6.35
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?
S 9.6.35
- \(H_{0}: p = 0.488\) \(H_{a}: p \neq 0.488\)
- \(p\text{-value} = 0.0114\)
- \(\alpha = 0.05\)
- Reject the null hypothesis.
- At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
- The survey does not appear to be accurate.
Q 9.6.36
Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \(\alpha = 0.05\), is the AAA proportion accurate?
Q 9.6.37
The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \(\alpha = 0.05\) level in Kentucky? Are the results applicable across the country? Why?
S 9.6.37
- \(H_{0}: p = 0.517\) \(H_{0}: p \neq 0.517\)
- \(p\text{-value} = 0.9203\).
- \(\alpha = 0.05\).
- Do not reject the null hypothesis.
- At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
- However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.
Q 9.6.38
For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use \(\alpha = 0.01\) level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?
Q 9.6.39
The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \(\alpha = 0.05 level\), can it be concluded that the mean rainfall was below the reported average? What if \(\alpha = 0.01\)? Assume the amount of summer rainfall follows a normal distribution.
S 9.6.39
- \(H_{0}: \mu \geq 11.52\) \(H_{a}: \mu < 11.52\)
- \(p\text{-value} = 0.000002\) which is almost 0.
- \(\alpha = 0.05\).
- Reject the null hypothesis.
- At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
- We would make the same conclusion if alpha was 1% because the \(p\text{-value}\) is almost 0.
Q 9.6.40
A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \(\alpha = 0.10\) level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?
Q 9.6.41
A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals
3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1
At the \(\alpha = 0.05\) level can it be concluded that the sample mean is higher than 5.8 visits per year?
S 9.6.42
- \(H_{0}: \mu \leq 5.8\) \(H_{a}: \mu > 5.8\)
- \(p\text{-value} = 0.9987\)
- \(\alpha = 0.05\)
- Do not reject the null hypothesis.
- At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.
Q 9.6.42
According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:
5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2
At \(\alpha = 0.05\) level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?
Q 9.6.43
The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?
S 9.6.43
- \(H_{0}: \mu \geq 150\) \(H_{0}: \mu < 150\)
- \(p\text{-value} = 0.0622\)
- \(\alpha = 0.01\)
- Do not reject the null hypothesis.
- At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
- The student academic group’s claim appears to be correct.


1.2: OSHA Procedures
OSHA Procedures
OSHA has set procedures for conducting worksite inspections, assessing civil penalties for compliance violations, delegating employer and employee responsibilities, filing complaints, promoting changes and rulemaking, codifying requirements, recordkeeping and reporting.
OSHA Inspections
To enforce its standards, OSHA has the authority to conduct inspections. Every place of employment covered by OSHA is subject to inspection. OSHA has the authority to enter without delay, at reasonable times, to inspect and investigate any place of employment. Inspections are conducted without advance notice. If an employee refuses to admit an OSHA compliance officer, OSHA will pursue legal actions, such as obtaining a search warrant.
Inspection priorities
Because of the vast amount of jobsites and the shortage of OSHA compliance personnel, OSHA has inspection priorities. The first priority is for imminent danger conditions. These are situations where there is reasonable certainty that a danger exists which can be expected to cause or is causing serious physical harm or death. The second priority is given to jobsite catastrophes and fatalities. The third priority is given to investigating employee complaints and the last priority is for programmed high-hazard inspections. OSHA also includes as part of the last priority, follow-up visits for compliance with abatement and mitigation of citations.
Inspection procedures
When an inspection occurs, OSHA has a set of procedures it follows. After the OSHA inspector shows his or her credentials and announces the inspection, an Opening Conference is held with the employer to explain the purpose of the visit, the scope of the inspection and the standards that apply. An employer representative and an employee representative are permitted to attend the conference and participate in the inspection. The Inspection Tour is the second step of the inspection. The OSHA officer and accompanying representatives proceed through the jobsite on the inspection. OSHA compliance officers are permitted to question employees during the tour and they will point out unsafe or unhealthy working conditions observed. The last step of the visit is the Closing Conference. The OSHA officer will discuss what was observed during the inspection and indicate all apparent violations for which a citation may be issued. The employer is told of his/her appeal rights.
No discussion of any proposed fines should occur at the conference. The OSHA area director is responsible for that determination only after having received a full report.
Penalty schedule
OSHA has established a penalty schedule based on the severity of the violation. Violations that are not likely to affect health and safety are classified as Other Than Serious and are subject to a proposed penalty of up to $13,653 for each violation. Violations where there is a probability that death or serious physical harm could occur are classified as Serious. A mandatory fine of $13,653 for each penalty is proposed. Willful violations occur when the employer intentionally and knowingly permits a hazardous condition to exist or makes no reasonable effort to abate such a hazard. Penalties of up to $136,532 per violation are proposed for willful violations. OSHA also has fines for Repeat violations ($136,532) and Failure to Correct Prior Violations ($13,653).
Employer Responsibilities
All employers have the primary responsibility to meet the provisions of the general duty clause, section 5(a)(1), by providing their employees a workplace that is free from recognized hazards that are causing or are likely to cause death or serious physical harm to their employees.
Employers must also comply with any standards, rules and regulations issued by OSHA. Employers must examine the workplace to make sure that all workplace conditions conform to the applicable standards. Employers must provide to their employees all of the necessary training required by OSHA standards.
Employee Responsibilities
While OSHA does not cite employees, employees have the responsibility to comply with all OSHA standards and all rules, regulations and orders issued by the OSHA Act. Every employee should follow the safety and health rules and regulations and wear appropriate Personal Protective Equipment (PPE) where so required. Employees should report any job related injury or illness to the employer immediately.
Employees’ Rights
Employees should always exercise their rights under the OSHA Act in a responsible manner. If an employee is exercising these or any other OSHA rights, the employer is not permitted to discriminate against that worker in any way, such as through firing, demotion, taking away benefits, transferring the worker to an undesirable job or shift, or threatening or harassing the worker.
Workers who believe that they have been punished or discriminated against for exercising their safety and health rights must contact the nearest OSHA office within 30 days of the time they learn of the alleged discrimination.
A union representative can file a complaint on behalf of the worker. The worker does not have to complete any forms. If necessary, OSHA will pursue legal action against the employer and the employee does not have to pay any legal fees.
OSHA has a 24-Hour Emergency Service Hotline for those who want to contact OSHA about life-threatening workplace hazards or serious health emergencies. 1-(800)-321-OSHA.
Development of OSHA regulations
OSHA regulations can be developed by OSHA itself, the Secretary of Health and Human Services, the National Institute for Occupational Safety and Health, state and local governments, nationally-recognized standards-producing organizations, employer or labor representatives or any other interested persons.


26.1: Introduction (2017)
Herman Melville was born in 1819 in New York City. He caught the scarlet fever around the age of five, and though he recovered quickly, his vision was permanently impaired. His father, Allan Melville, was a high-end importer and merchant. The Melvilles lived a prosperous life until Allan’s sudden death in 1832. To support himself, Melville worked as a bank clerk in the 1830s. At the same time, he was enrolled at Albany Classical School where he studied classic literature and began to write essays, poems and short stories. Once out of school, Melville tried his hand at a number of jobs including school teaching and surveying.
Melville eventually landed a job as a cabin boy in 1839 on a merchant ship. Satisfied, he went on his second voyage in 1841 on a whaling ship named the Acushnet. This second voyage ended up being the inspiration for Melville’s future writing career. The Acushnet brought Melville down to the Polynesian islands in 1842. Soon after, he and another crewmember deserted the ship and were captured by cannibals. Melville eventually escaped the islands four months later on another whaling ship named the Lucy Ann, but was locked up for joining the crew in an attempted mutiny. He eventually wound up in Hawaii where he made his way back to Massachusetts on the USS United States; he had been gone nearly four years.
Melville chose to write about this instance of slavery because of the Tryal Slave Ship Rebellion of 1805. He started writing the story after reading Captain Amasa Delano’s memoirs. It started as a slave rising on ship in the South Pacific near Chile. West Africans were bought by the Spanish and they were put on the ship to be brought to Buenos Aires in Argentina. At the time Argentina was a Spanish colony. When they arrived, they were brought to the port of Valpariso and the 72 Africans that survived boarded the Spanish ship, Tryal. The ship was meant to bring the slaves to the north slave markets in Peru and Ecuador.
In 1804, on their way there; Babo and Mori, the two leaders of the enslaved Africans, lead a successful rebellion. They took control and killed most of the Spanish aboard the ship but kept Captain Benito Cereno as a hostage. In February 1805, Captain Amasa Delano came across the Tryal ship. Delano brought them food and water but when he went back to his own Ship, Benito Cereno broke free and Delano’s crew members rushed back on the Tryal ship and killed most of the Africans.  The Africans that remained were brought to Spanish authorities. Melville took what he had learned from the trial and of Delano’s memoirs and turned it into his own literary work.
Due to the ambiguity of “Benito Cereno” (1856), readers often find it challenging to determine what stance—if any—the text takes on slavery. Written just before American Civil War, Melville’s novella neither supports nor denounces slavery explicitly. However, many find the text’s focus on the topic to be part of a larger theme regarding the meaning of innocence. Melville’s choice to tell the story from a point of view that limits readers to a fictionalized version Captain Delano’s experiences often evokes a controversial sense of sympathy for all parties aboard the San Dominick.
The point of view in Melville’s text also caters to the theme of American identity. According to scholars such as Paul Downes, the story’s perspective “produced a character whose naïveté … laid bare the lines of association and identification that tie vulnerability to power, revolution to violence, and citizenship to rightlessness” (479). In other words, readers like Downes believe that Melville’s use of deception and naivety make the challenges within American ideology clearer. Delano, the only American in “Benito Cereno,” is portrayed as a strong, generous, and intelligent captain. Yet his inability to accurately assess the situation on the San Dominick challenges these portrayals and causes many to perceive his naivety as a nod to unsupported American exceptionalism.
The undying desire to define the American identity has kept “Benito Cereno” a canonized piece of literature throughout the years. While the situations of today differ from those of the 19th century, the story’s themes of innocence, self-identity, and the power (and danger) of naivety are consistently relevant within American society. This ambiguity and timeless relevance allows scholars to continue turning back to Melville’s novella for clues as to what it means to be an American.
Works Cited:
Anderson, John. “The Tryal Slave Ship Rebellion, 1805.” BlackPast.org, n.d., blackpast.org/gah/tryal-slave-ship-rebellion-1805.
Biography.com Editors. “Herman Melville.” Biography.com, A&E Networks Television, 9 Dec. 2015, biography.com/people/herman-melville-9405239.
Downes, Paul. “Melville’s Benito Cereno and the Politics of Humanitarian Intervention.”
The South Atlantic Quarterly
, vol. 103 no. 2, 2004, pp. 465-488. Project MUSE, muse.jhu.edu/article/169143.
Melville, Herman. “Benito Cereno.” 1856.


22.4: Chapter Summary
Transcultural nursing involves providing client-centered care that is culturally competent and sensitive to individuals from diverse cultural backgrounds. The practice of transcultural nursing is based on understanding culture’s influence on health care decisions, illness, and health. Transcultural nursing aims to promote culturally inclusive health care, enhance client engagement, and reduce health care disparities by incorporating the cultural beliefs, values, and practices of clients into the plan of care.
Cultural models are frameworks or patterns of thinking and behavior shared among members of a particular cultural group. These models influence how individuals perceive and interpret the world, including their views on health, wellness, and health care. Nurses can use these models to effectively communicate, establish trust, and tailor interventions that align with clients’ cultural beliefs and values, ultimately improving client outcomes and satisfaction.
Cultural assessment in nursing is the systematic process of gathering information about clients’ cultural beliefs, values, norms, and practices to better understand their health care needs. It involves exploring the client’s cultural background, language preferences, religious or spiritual beliefs, family dynamics, and health care practices to provide culturally competent care. Through cultural assessment, nurses can identify any cultural barriers or influences that may impact client care and develop individualized care plans that respect and accommodate the client’s cultural context, promoting better health outcomes and client satisfaction.


6.4.1: The top 10 leading causes of death for Americans in 2023
- Heart disease
- Cancers
- Unintentional injuries
- Stroke
- Chronic lower respiratory diseases
- Alzheimer disease
- Diabetes
- Kidney disease
- Chronic liver disease and cirrhosis
- COVID-19
Deaths at younger ages may be more concerning from a public health standpoint, since many of these deaths are below the average lifespan typically expected, and therefore they may be preventable. Those years lost from an early death include a loss of physical, emotional, and financial support to that individual’s family, and friends. That persons passing leaves a hole in the social fabric of their community. And, although it may seem trite to mention, there is also a significant economic impact from early deaths. The years of potential life lost or YPLL is another calculation used by epidemiology to express the number of deaths at younger ages, and the impact that those younger deaths have on the population. This YPLL is the difference between the age of deaths and a “standard year” of death - which is often set somewhere between 65-75 years of age. For example using the higher number, a child’s death at 5 years of age would have a YPLL of 70, whereas an older adult’s death at age 65 would have a YPLL of 10. Deaths above the standard year are not counted. It’s important to know the standard year used for this calculation, and why that particular number was chosen. This metric can be used for a specific disease, condition or a population. This calculation can be used to help focus public health efforts on preventing diseases with the greatest impact to younger populations (Seabert et al., 2021).
Sometimes the YPLL can also provide a more complete picture of the impact of a disease or condition than just the mortality rate alone. A great example of this is not a disease at all, but rather fatal accidents and violence - which tend to impact younger age groups the most heavily. Unintentional injuries (including car accidents and drug overdoses), homicide and suicide are leading causes of death for youths between 15-19 years of age ( FastStats , 2023b). If we look at the data for unintentional injuries in 2021 overall, there were 224,935 deaths due to unintentional injuries. This equates to a crude (unadjusted) rate of 67.77 per 100,000 people, and an age-adjusted rate of 64.67 per 100,000. The age adjustment doesn’t have a large effect on the mortality rate, likely due to the fact that unintentional injuries are a leading cause of death for younger age ranges - that is, there isn’t an increasing risk of death with advancing age, like there may be with another cause of death like heart disease (which is a chronic disease, and therefore risk of death increases with age). The CDC calculated the number of years of potential life lost in 2021 due to unintentional injuries at 3,781,756 - which demonstrates the significant impact that this cause of death has on American society ( WISQARS Fatal and Nonfatal Injury Reports , n.d.).
Still other metrics attempt to quantify the impact of chronic disease and disability on populations. Sometimes the term healthy life expectancy (HALE) is used to determine how many average years a person can expect to live before experiencing debilitating effects of chronic disease. This number varies depending on the age we begin counting from, so often HALE at birth is used to determine a lifespan estimate and HALE60 - or healthy life expectancy at 60 years is used to describe the burden of disease and disability on our current aging population. For example, according to data from the World Health Organization, HALE60 in the U.S. in 2019 was 16.35 years - so someone at age 60 in 2019 could have expected to live on average another 16 or so years without disability. In the same year, Canada’s HALE 60 was 18.99 years - demonstrating that Canada has a longer HALE60, and perhaps slightly lower burden of disease on their aging population (World Health Organization, 2023b).
Years lost to early death or chronic disease are also sometimes combined with years of life under which a person experiences life-altering disability. This metric is termed disability adjusted life years (DALYs), where one DALY equals a year of healthy life lost either to premature death or to disability. The World Health Organization has estimated the top diseases and conditions which are the highest sources of DALYs for countries around the world. In 2019, the leading causes of years lost to premature death and disability included (in order) (World Health Organization, 2023a):
- Ischemic heart disease
- Drug use disorders
- Chronic obstructive pulmonary disease
- Diabetes mellitus
- Neck and back pain
- Alzheimer disease and dementias
- Stroke
- Cancers of the lungs, trachea and bronchi
- Depressive disorders
- Self-harm
This list demonstrates that the leading causes of disability and premature death may not always be the same as the leading causes of death. Using these other metrics like YPLL and DALY allows for epidemiologists to get a fuller picture of not only how long a population is living, but how healthy those years are, and the causes of harm experienced.


The group project was made of two parts—Research and Presentation.
Research
By the third week of the semester, participants in the course were assigned to a global collaborative research team called Global Virtual Teams.6 Each Global Virtual Team consisted of five persons (four from KUC and one from UGL.) Each Virtual Team was assigned one of five stakeholder perspectives:
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global private sector;
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international organization;
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developed country national government;
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developing country national government; and
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non-governmental organization (non-state actors or NGO).
These Global Virtual Teams were tasked with a research problem and a role-playing exercise. Each global virtual team was expected to develop a 4,000–5,000 word e-essay/paper and a 15-minute (Address to Humanity) presentation on the following research questions:
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What is Globalization?
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Why has it attracted much controversy, supporters and detractors?
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How has globalization contributed to the wealth and poverty of nations? Identify the problems and promises of globalization.
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What roles should governments, individuals, civic society, the UN play in this?
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Propose three ways in which valued resources such as energy, food, shelter, medicine, etc., can be equitably and justly distributed.
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The paper must be based on one of the areas to be covered in the course listed below.
Presentation
Fifteen-Minute Address to Humanity: Mock UN Assembly Meeting: The Global Virtual Teams were expected to present a summary of their paper to an imaginary United Nations session devoted to Globalization. This was done during the four weeks of the term/ semester.
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Culture
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Social Justice
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Economic Development
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Indigenous Peoples
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Foreign Policy
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Global Climate
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Global Health
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International Conflict
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Democracy
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Migration
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Religion
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Trade
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The Media
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Women
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Children
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Human Rights
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Racial/Ethnic Minorities
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Senior Citizens
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7.17: Connections of Carbohydrate, Protein, and Lipid Metabolic Pathways - Connecting Lipids to Glucose Metabolism
- Explain the connection of lipids to glucose metabolism
Like sugars and amino acids, the catabolic pathways of lipids are also connected to the glucose catabolism pathways. The lipids that are connected to the glucose pathways are cholesterol and triglycerides.
Cholesterol
Cholesterol contributes to cell membrane flexibility and is a precursor to steroid hormones. The synthesis of cholesterol starts with acetyl groups, which are transferred from acetyl CoA, and proceeds in only one direction; the process cannot be reversed. Thus, synthesis of cholesterol requires an intermediate of glucose metabolism.
Triglycerides
Triglycerides, a form of long-term energy storage in animals, are made of glycerol and three fatty acids. Animals can make most of the fatty acids they need. Triglycerides can be both made and broken down through parts of the glucose catabolism pathways. Glycerol can be phosphorylated to glycerol-3-phosphate, which continues through glycolysis.
Fatty acids are catabolized in a process called beta-oxidation that takes place in the matrix of the mitochondria and converts their fatty acid chains into two carbon units of acetyl groups, while producing NADH and FADH 2 . The acetyl groups are picked up by CoA to form acetyl CoA that proceeds into the citric acid cycle as it combines with oxaloacetate. The NADH and FADH 2 are then used by the electron transport chain.
Contributions and Attributions
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44441/latest...ol11448/latest . License : CC BY: Attribution
- Medical Physiology/Basic Biochemistry/Sugars. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Medical...ose_Metabolism . License : CC BY-SA: Attribution-ShareAlike
- glycogen. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/glycogen . License : CC BY-SA: Attribution-ShareAlike
- monosaccharide. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/monosaccharide . License : CC BY-SA: Attribution-ShareAlike
- disaccharide. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/disaccharide . License : CC BY-SA: Attribution-ShareAlike
- Medical Physiology/Basic Biochemistry/Sugars. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Medical...ose_Metabolism . License : CC BY-SA: Attribution-ShareAlike
- Glycogen Structure. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Glycog..._structure.svg . License : Public Domain: No Known Copyright
- Glycogen pathway.jpg. Provided by : OpenStax CNX. Located at<EMAIL_ADDRESS>. License : CC BY-SA: Attribution-ShareAlike
- keto acid. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/keto_acid . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44441/latest...ol11448/latest . License : CC BY: Attribution
- deamination. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/deamination . License : CC BY-SA: Attribution-ShareAlike
- catabolism. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/catabolism . License : CC BY-SA: Attribution-ShareAlike
- Medical Physiology/Basic Biochemistry/Sugars. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Medical_Physiology/Basic_Biochemistry/Sugars%23Galactose_Metabolism . License : CC BY-SA: Attribution-ShareAlike
- Glycogen Structure. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Glycogen%23mediaviewer/File:Glycogen_structure.svg . License : Public Domain: No Known Copyright
- Glycogen pathway.jpg. Provided by : OpenStax CNX. Located at : https://cnx.org/contents/ZP457F64@7/Connections-of-Carbohydrate-Pr . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Connections of Carbohydrate, Protein, and Lipid Metabolic Pathways. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44441/latest...e_07_06_01.jpg . License : CC BY: Attribution
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44441/latest/?collection=col11448/latest . License : CC BY: Attribution
- Metabolomics/Metabolites/Lipids/Energy Storage. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Metabol...Energy_Storage . License : CC BY-SA: Attribution-ShareAlike
- Boundless. Provided by : Boundless Learning. Located at : www.boundless.com//biology/de...beta-oxidation . License : CC BY-SA: Attribution-ShareAlike
- lipid. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/lipid . License : CC BY-SA: Attribution-ShareAlike
- Medical Physiology/Basic Biochemistry/Sugars. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Medical_Physiology/Basic_Biochemistry/Sugars%23Galactose_Metabolism . License : CC BY-SA: Attribution-ShareAlike
- Glycogen Structure. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Glycogen%23mediaviewer/File:Glycogen_structure.svg . License : Public Domain: No Known Copyright
- Glycogen pathway.jpg. Provided by : OpenStax CNX. Located at : https://cnx.org/contents/ZP457F64@7/Connections-of-Carbohydrate-Pr . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Connections of Carbohydrate, Protein, and Lipid Metabolic Pathways. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44441/latest...e_07_06_01.jpg . License : CC BY: Attribution
Key Points
- Many types of lipids exist, but cholesterol and triglycerides are the lipids that enter the pathways of glucose catabolism.
- Through the process of phosphorylation, glycerol can be converted to glycerol-3-phosphate during the glycolytic pathway.
- When fatty acids are broken down into acetyl groups through beta-oxidation, the acetyl groups are used by CoA to form acetyl-CoA, which enters the citric acid cycle to produce ATP.
- Beta-oxidation produces FADH 2 and NADH, which are used by the electron transport chain for ATP production.
Key Terms
- beta-oxidation : A process that takes place in the matrix of the mitochondria and catabolizes fatty acids by converting them to acetyl groups while producing NADH and FADH2.
- lipid : A group of organic compounds including fats, oils, waxes, sterols, and triglycerides; characterized by being insoluble in water; account for most of the fat present in the human body.


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4.1: Magnetism
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The magnetic moment of a material is the incomplete cancellation of the atomic magnetic moments in that material. Electron spin and orbital motion both have magnetic moments associated with them but in most atoms the electronic moments are oriented usually randomly so that overall in the material they cancel each other out; this is called diamagnetism.
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4.2: IR Spectroscopy
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Infrared spectroscopy is based on molecular vibrations caused by the oscillation of molecular dipoles. Bonds have characteristic vibrations depending on the atoms in the bond, the number of bonds and the orientation of those bonds with respect to the rest of the molecule. Thus, different molecules have specific spectra that can be collected for use in distinguishing products or identifying an unknown substance (to an extent.)
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4.3: Raman Spectroscopy
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Raman spectroscopy is a powerful tool for determining chemical species. As with other spectroscopic techniques, Raman spectroscopy detects certain interactions of light with matter. In particular, this technique exploits the existence of Stokes and Anti-Stokes scattering to examine molecular structure.
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4.4: UV-Visible Spectroscopy
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Ultraviolet-visible (UV-vis) spectroscopy is used to obtain the absorbance spectra of a compound in solution or as a solid. What is actually being observed spectroscopically is the absorbance of light energy or electromagnetic radiation, which excites electrons from the ground state to the first singlet excited state of the compound or material. The UV-vis region of energy for the electromagnetic spectrum covers 1.5 .2 eV which relates to a wavelength range of 800 - 200 nm.
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4.5: Photoluminescence, Phosphorescence, and Fluorescence Spectroscopy
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Photoluminescence spectroscopy is a contactless, nondestructive method of probing the electronic structure of materials. Light is directed onto a sample, where it is absorbed and imparts excess energy into the material in a process called photo-excitation. One way this excess energy can be dissipated by the sample is through the emission of light, or luminescence. In the case of photo-excitation, this luminescence is called photoluminescence.
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4.6: Mössbauer Spectroscopy
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In 1957 Rudolf Mossbauer achieved the first experimental observation of the resonant absorption and recoil-free emission of nuclear γ-rays in solids during his graduate work at the Institute for Physics of the Max Planck Institute for Medical Research in Heidelberg Germany. Mossbauer received the 1961 Nobel Prize in Physics for his research in resonant absorption of γ-radiation and the discovery of recoil-free emission a phenomenon that is named after him. The Mossbauer effect is the basis of Mo
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4.7: NMR Spectroscopy
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Nuclear magnetic resonance spectroscopy (NMR) is a widely used and powerful method that takes advantage of the magnetic properties of certain nuclei. The basic principle behind NMR is that some nuclei exist in specific nuclear spin states when exposed to an external magnetic field.
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4.8: EPR Spectroscopy
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Electron paramagnetic resonance spectroscopy (EPR) is a powerful tool for investigating paramagnetic species, including organic radicals, inorganic radicals, and triplet states. The basic principles behind EPR are very similar to the more ubiquitous nuclear magnetic resonance spectroscopy (NMR), except that EPR focuses on the interaction of an external magnetic field with the unpaired electron(s) in a molecule, rather than the nuclei of individual atoms.
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4.9: X-ray Photoelectron Spectroscopy
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X-ray photoelectron spectroscopy (XPS), also called electron spectroscopy for chemical analysis (ESCA), is a method used to determine the elemental composition of a material’s surface. It can be further applied to determine the chemical or electronic state of these elements.
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4.10: ESI-QTOF-MS Coupled to HPLC and its Application for Food Safety
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Mass spectrometry (MS) is a detection technique by measuring mass-to-charge ratio of ionic species. The procedure consists of different steps. First, a sample is injected in the instrument and then evaporated. Second, species in the sample are charged by certain ionized methods, such as electron ionization (EI), electrospray ionization (ESI), chemical ionization (CI), matrix-assisted laser desorption/ionization (MALDI).
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4.11: Mass Spectrometry
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Mass spectrometry (MS) is a powerful characterization technique used for the identification of a wide variety of chemical compounds. At its simplest, MS is merely a tool for determining the molecular weight of the chemical species in a sample. However, with the high resolution obtainable from modern machines, it is possible to distinguish isomers, isotopes, and even compounds with nominally identical molecular weights. Libraries of mass spectra have been compiled which allow rapid identification
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10.1: Introduction
The Sierra Nevada is a mountain range in the Western United States, between the Central Valley of California and the Great Basin. Most of the range lies in the state of California, although the Carson Range spur lies primarily in Nevada. The Sierra Nevada is part of the American Cordillera, an almost continuous chain of mountain ranges that forms the western "backbone" of the Americas.
The Sierra runs 400 miles (640 km) north-south and its width ranges from 50 miles (80 km) to 80 miles (130 km) across east–west. Notable features include General Sherman, the largest tree in the world by volume; Lake Tahoe, the largest alpine lake in North America; Mount Whitney at 14,505 ft (4,421 m), the highest point in the contiguous United States; and Yosemite Valley sculpted by glaciers from one-hundred-million-year-old granite, containing high waterfalls. The Sierra is home to three national parks, twenty-six wilderness areas, ten national forests, and two national monuments. These areas include Yosemite, Sequoia, and Kings Canyon National Parks; and Devils Postpile National Monument.
More than one hundred million years ago during the Nevadan orogeny, granite formed deep underground. The range started to uplift less than five million years ago, and erosion by glaciers exposed the granite and formed the light-colored mountains and cliffs that make up the range. The uplift caused a wide range of elevations and climates in the Sierra Nevada, which are reflected by the presence of five life zones (areas with similar plant and animal communities). Uplift continues due to faulting caused by tectonic forces, creating spectacular fault block escarpments along the eastern edge of the southern Sierra.
The Sierra Nevada has played an important role in the history of California and the United States. The California Gold Rush occurred in the western foothills from 1848 through 1855. Due to its inaccessibility, the range was not fully explored until 1912.


3.3: Linear Equations
The linear first-order differential equation (linear in \(y\) and its derivative) can be written in the form \[\label{eq:1}\frac{dy}{dx}+p(x)y=g(x),\] with the initial condition \(y(x_0) = y_0\). Linear first-order equations can be integrated using an integrating factor \(\mu(x)\). We multiply \(\eqref{eq:1}\) by \(\mu(x)\), \[\label{eq:2}\mu (x)\left[\frac{dy}{dx}+p(x)y\right]=\mu (x)g(x),\] and try to determine \(\mu (x)\) so that \[\label{eq:3}\mu (x)\left[\frac{dy}{dx}+p(x)y\right]=\frac{d}{dx}[\mu (x)y].\]
Equation \(\eqref{eq:2}\) then becomes \[\label{eq:4}\frac{d}{dx}[\mu (x)y]=\mu (x)g(x).\]
Equation \(\eqref{eq:4}\) is easily integrated using \(\mu (x_0)=\mu_0\) and \(y(x_0)=y_0\):
\[\mu (x)y-\mu_0 y_0=\int_{x_0}^x\mu (x)g(x)dx,\nonumber\] or \[\label{eq:5}y=\frac{1}{\mu (x)}\left(\mu_0y_0+\int_{x_0}^x\mu(x)g(x)dx\right).\]
It remains to determine \(\mu(x)\) from \(\eqref{eq:3}\). Differentiating and expanding \(\eqref{eq:3}\) yields \[\mu\frac{dy}{dx}+p\mu y=\frac{d\mu}{dx}y+\mu\frac{dy}{dx};\nonumber\] and upon simplifying, \[\label{eq:6}\frac{d\mu}{dx}=p\mu.\]
Equation \(\eqref{eq:6}\) is separable and can be integrated:
\[\begin{aligned}\int_{\mu_0}^{\mu}\frac{d\mu}{\mu}&=\int_{x_0}^xp(x)dx, \\ \ln\frac{\mu}{\mu_0}&=\int_{x_0}^xp(x)dx, \\ \mu(x)&=\mu_0\exp\left(\int_{x_0}^xp(x)dx\right).\end{aligned}\]
Notice that since \(\mu_0\) cancels out of \(\eqref{eq:5}\), it is customary to assign \(\mu_0 = 1\). The solution to \(\eqref{eq:1}\) satisfying the initial condition \(y(x_0) = y_0\) is then commonly written as \[y=\frac{1}{\mu(x)}\left(y_0+\int_{x_0}^x\mu(x)g(x)dx\right),\nonumber\] with \[\mu(x)=\exp\left(\int_{x_0}^xp(x)dx\right)\nonumber\] the integrating factor. This important result finds frequent use in applied mathematics.
Solve \(\frac{dy}{dx}+2y=e^{-x}\), with \(y(0)=3/4\).
Solution
Note that this equation is not separable. With \(p(x) = 2\) and \(g(x) = e^{−x}\), we have \[\begin{aligned}\mu(x)&=\exp\left(\int_0^x 2dx\right) \\ &=e^{2x},\end{aligned}\] and \[\begin{aligned}y&=e^{-2x}\left(\frac{3}{4}+\int_0^x e^{2x}e^{-x}dx\right) \\ &=e^{-2x}\left(\frac{3}{4}+\int_0^x e^xdx\right) \\ &=e^{-2x}\left(\frac{3}{4}+(e^x-1)\right) \\ &=e^{-2x}\left(e^x-\frac{1}{4}\right) \\ &=e^{-x}\left(1-\frac{1}{4}e^{-x}\right).\end{aligned}\]
Solve \(\frac{dy}{dx}-2xy=x\), with \(y(0)=0\).
Solution
This equation is separable, and we solve it in two ways. First, using an integrating factor with \(p(x) = −2x\) and \(g(x) = x\):
\[\begin{aligned}\mu(x)&=\exp\left(-2\int_0^x xdx\right) \\ &=e^{-x^2},\end{aligned}\] and \[y=e^{x^2}\int_0^x xe^{-x^2}dx.\nonumber\]
The integral can be done by substitution with \(u = x^2\), \(du = 2xdx\):
\[\begin{aligned}\int_0^x xe^{-x^2}dx&=\frac{1}{2}\int_0^{x^2}e^{-u}du \\ &=-\frac{1}{2}e^{-u}]_0^{x^2}\\ &=\frac{1}{2}\left(1-e^{-x^2}\right).\end{aligned}\]
Therefore, \[\begin{aligned}y&=\frac{1}{2}e^{x^2}\left(1-e^{-x^2}\right) \\ &=\frac{1}{2}\left(e^{x^2}-1\right).\end{aligned}\]
Second, we integrate by separating variables:
\[\begin{aligned}\frac{dy}{dx}-2xy&=x, \\ \frac{dy}{dx}&=x(1+2y), \\ \int_0^y\frac{dy}{1+2y}&=\int_0^x xdx, \\ \frac{1}{2}\ln (1+2y)&=\frac{1}{2}x^2, \\ 1+2y&=e^{x^2}, \\ y&=\frac{1}{2}\left(e^{x^2}-1\right).\end{aligned}\]
The results from the two different solution methods are the same, and the choice of method is a personal preference.


2.2: Respiratory Structures and Their Function
- Describe select respiratory structures in aquatic and terrestrial animals
- Explain relationships between respiratory structure and function in generalized aquatic and terrestrial animal systems.
Skin and Gills
Earthworms and amphibians use their skin (integument) as a respiratory organ. A dense network of capillaries lies just below the skin and facilitates gas exchange between the external environment and the circulatory system. The respiratory surface must be kept moist in order for the gases to dissolve and diffuse across cell membranes.
Organisms that live in water need to obtain oxygen from the water. Oxygen dissolves in water but at a lower concentration than in the atmosphere. The atmosphere has roughly 21 percent oxygen. In water, the oxygen concentration is much smaller than that. Fish and many other aquatic organisms have evolved gills to take up the dissolved oxygen from water (Figure 2.4). Gills are thin tissue filaments that are highly branched and folded. When water passes over the gills, the dissolved oxygen in water rapidly diffuses across the gills into the bloodstream. The circulatory system can then carry the oxygenated blood to the other parts of the body. In animals that contain coelomic fluid instead of blood, oxygen diffuses across the gill surfaces into the coelomic fluid. Gills are found in mollusks, annelids, and crustaceans.
The folded surfaces of the gills provide a large surface area to ensure that the fish gets sufficient oxygen. Diffusion is a process in which material travels from regions of high concentration to low concentration until equilibrium is reached. In this case, blood with a low concentration of oxygen molecules circulates through the gills. The concentration of oxygen molecules in water is higher than the concentration of oxygen molecules in gills. As a result, oxygen molecules diffuse from water (high concentration) to blood (low concentration), as shown in Figure 2.5 . Similarly, carbon dioxide molecules in the blood diffuse from the blood (high concentration) to water (low concentration).
Tracheal systems
Insect respiration is independent of its circulatory system; therefore, the blood does not play a direct role in oxygen transport. Insects have a highly specialized type of respiratory system called the tracheal system, which consists of a network of small tubes that carries oxygen to the entire body. The tracheal system is the most direct and efficient respiratory system in active animals. The tubes in the tracheal system are made of a polymeric material called chitin.
Insect bodies have openings, called spiracles along the thorax and abdomen. These openings connect to the tubular network, allowing oxygen to pass into the body (Figure 2.6) and regulating the diffusion of CO 2 and water vapor. Air enters and leaves the tracheal system through the spiracles. Some insects can ventilate the tracheal system with body movements.
Mammalian systems
In mammals, pulmonary ventilation occurs via inhalation (breathing). During inhalation, air enters the body through the nasal cavity located just inside the nose (Figure 2.7). As air passes through the nasal cavity, the air is warmed to body temperature and humidified. The respiratory tract is coated with mucus to seal the tissues from direct contact with air. Mucus is high in water. As air crosses these surfaces of the mucous membranes, it picks up water. These processes help equilibrate the air to the body conditions, reducing any damage that cold, dry air can cause. Particulate matter that is floating in the air is removed in the nasal passages via mucus and cilia. The processes of warming, humidifying, and removing particles are important protective mechanisms that prevent damage to the trachea and lungs. Thus, inhalation serves several purposes in addition to bringing oxygen into the respiratory system.
Which of the following statements about the mammalian respiratory system is false?
- When we breathe in, air travels from the pharynx to the trachea.
- The bronchioles branch into bronchi.
- Alveolar ducts connect to alveolar sacs.
- Gas exchange between the lung and blood takes place in the alveolus
From the nasal cavity, air passes through the pharynx (throat) and the larynx (voice box), as it makes its way to the trachea (Figure 2.8). The main function of the trachea is to funnel the inhaled air to the lungs and the exhaled air back out of the body. The human trachea is a cylinder about 10 to 12 cm long and 2 cm in diameter that sits in front of the esophagus and extends from the larynx into the chest cavity where it divides into the two primary bronchi at the mid-thorax. It is made of incomplete rings of hyaline cartilage and smooth muscle (Figure 2.8). The trachea is lined with mucus-producing goblet cells and ciliated epithelia. The cilia propel foreign particles trapped in the mucus toward the pharynx. The cartilage provides strength and support to the trachea to keep the passage open. The smooth muscle can contract, decreasing the trachea’s diameter, which causes expired air to rush upwards from the lungs at a great force. The forced exhalation helps to expel mucus when we cough. Smooth muscle can contract or relax, depending on stimuli from the external environment or the body’s nervous system.
Lungs: bronchi and alveoli
The end of the trachea bifurcates (divides) to the right and left lungs. The lungs are not identical. The right lung is larger and contains three lobes, whereas the smaller left lung contains two lobes (Figure 2.9). The muscular diaphragm, which facilitates breathing, is inferior (below) to the lungs and marks the end of the thoracic cavity.
In the lungs, air is diverted into smaller and smaller passages or bronchi. Air enters the lungs through the two primary (main) bronchi (singular: bronchus). Each bronchus divides into secondary bronchi, then into tertiary bronchi, which in turn divide, creating smaller and smaller diameter bronchioles as they split and spread through the lung. Like the trachea, the bronchi are made of cartilage and smooth muscle. At the bronchioles, the cartilage is replaced with elastic fibers. Bronchi are innervated by nerves of both the parasympathetic and sympathetic nervous systems that control muscle contraction (parasympathetic) or relaxation (sympathetic) in the bronchi and bronchioles, depending on the nervous system’s cues. In humans, bronchioles with a diameter smaller than 0.5 mm are the respiratory bronchioles . They lack cartilage and therefore rely on inhaled air to support their shape. As the passageways decrease in diameter, the relative amount of smooth muscle increases.
The terminal bronchioles subdivide into microscopic branches called respiratory bronchioles. The respiratory bronchioles subdivide into several alveolar ducts. Numerous alveoli and alveolar sacs surround the alveolar ducts. The alveolar sacs resemble bunches of grapes tethered to the end of the bronchioles (Figure 2.10). In the acinar region, the alveolar ducts are attached to the end of each bronchiole. At the end of each duct are approximately 100 alveolar sacs, each containing 20 to 30 alveoli that are 200 to 300 microns in diameter. Gas exchange occurs only in alveoli. Alveoli are made of thin-walled parenchymal cells, typically one-cell thick, that look like tiny bubbles within the sacs. Alveoli are in direct contact with capillaries (one-cell thick) of the circulatory system. Such intimate contact ensures that oxygen will diffuse from alveoli into the blood and be distributed to the cells of the body. In addition, the carbon dioxide that was produced by cells as a waste product will diffuse from the blood into alveoli to be exhaled. The anatomical arrangement of capillaries and alveoli emphasizes the structural and functional relationship of the respiratory and circulatory systems. Because there are so many alveoli (~300 million per lung) within each alveolar sac and so many sacs at the end of each alveolar duct, the lungs have a sponge-like consistency. This organization produces a very large surface area that is available for gas exchange. The surface area of alveoli in the lungs is approximately 75 m2. This large surface area, combined with the thin-walled nature of the alveolar parenchymal cells, allows gases to easily diffuse across the cells.
Which of the following statements about the human respiratory system is false?
a. When we breathe in, air travels from the pharynx to the trachea.
b. The bronchioles branch into bronchi.
c. Alveolar ducts connect to alveolar sacs.
d. Gas exchange between the lungs and blood takes place in the alveolus


11: Energy and Chemical Processes
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Energy is a vital component of the world around us. Nearly every physical and chemical process, including all the chemical reactions discussed in previous chapters, occurs with a simultaneous energy change. In this chapter, we will explore the nature of energy and how energy and chemistry are related.
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11.1: Prelude to Energy and Chemical Processes
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Metabolism is the collective term for the chemical reactions that occur in cells and provide energy to keep cells alive. Some of the energy from metabolism is in the form of heat, and some animals use this heat to regulate their body temperatures. Such warm-blooded animals are called endotherms. In endotherms, problems with metabolism can lead to fluctuations in body temperature. When humans get sick, for instance, our body temperatures can rise higher than normal; we develop a fever.
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11.2: Energy and Its Units
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Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules.
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11.3: Heat
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Heat transfer is related to temperature change. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat.
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11.4: Phase Changes
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There is an energy change associated with any phase change.
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11.5: Bond Energies and Chemical Reactions
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Atoms are held together by a certain amount of energy called bond energy. Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.
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11.6: The Energy of Biochemical Reactions
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Energy to power the human body comes from chemical reactions.
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11.E: Energy and Chemical Processes (Exercises)
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Problems and Solutions to accompany the chapter.
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11.S: Energy and Chemical Processes (Summary)
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To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
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2.1: Introduction
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The land of Sumer , in today’s southern Iraq, was home to some of the largest early cities in human history. In one of these ancient settlements, Ur, a beautiful wooden box was laid in a royal tomb in about 2550 BCE (Figure 3.1). It measures roughly nine by twenty inches (a little bigger than a laptop) and is inlaid with elaborate mosaic figures and borders composed of bits of red limestone, lapis lazuli, and marine shell. This kind of specialized craftsmanship was a hallmark of societies that no longer depended on hunting and gathering for food but rather produced crops capable of sustaining large populations. In turn, they gained enough time and prosperity for some members to focus on artisanal crafts.
The box indicates at least three important things about the civilization that produced it. First, a highly skilled artisan constructed the box and created the mosaics, indicating the presence of specialization of labor. Second, the mosaics show someone who is presumably the king at the center of the top row, directing the soldiers below. These power dynamics suggest new social hierarchies. Finally, the soldiers all appear smaller in the scene than the king, symbolically reflecting their subordinate position and telling us that social stratification had come into existence. All these developments took place gradually over time, bringing slow but enduring change to the lives of the people in Ur and those who lived nearby. Similar changes occurred in the world’s other ancient cities.